Suppose $X \in (0,1)$, and that $X^n < (1-e^{n\delta})$ for $\delta>0$ and $n \in \mathbb{N}$, is it true $X < (1-e^{n\delta})^{1/n}$?

Question

Suppose $X \in (0,1)$, and that $X^n < (1-e^{n\delta})$ for $\delta>0$ and $n \in \mathbb{N}$, then, I am trying to see if

$$ X < (1-e^{n\delta})^{1/n} $$

is true? In other words, is the $n$th root a monotone function here?

Answer 1

The $n$-th root function is strictly increasing on the set $E$ of non-negative real numbers (note that said function may not exist elsewhere, for we don't know the parity of $n$).

To see why, say $a,b \in E$ with $a<b$. If it were true that $\sqrt[n]{a} \geqslant \sqrt[n]{b}$ then by the axioms of an ordered field we would get the contradiction $a = \sqrt[n]{a} \cdots \sqrt[n]{a} \geqslant \sqrt[n]{b} \cdots \sqrt[n]{b} = b$ (here there are $n$ factors in either product). Thus it must be that $\sqrt[n]{a} < \sqrt[n]{b}$.